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Choose coordinates such that $\vec v = (v, 0, 0)$. Then simply read off the components of the Lorentz transformation
$$A_0 \rightarrow \Lambda_0\, ^\mu A_\mu = \Lambda_0 \, ^0 A_0 + \Lambda_0 \, ^1 A_1 = \gamma A_0 - \beta \gamma A_1.$$ Now $A_0 = \phi$, and by our choice of coordinates $\vec A \cdot \vec v = (v,0,0)\cdot(A_1,A_2,A_3) = v A_1$, hence identify $\vec \beta \cdot \vec A = \beta A_1$ to obtain $$A_0 \rightarrow \gamma ( \phi - \vec \beta \cdot \vec A)$$ as required.


Similarly for the components of $\vec A$
$$A_1 \rightarrow \Lambda_1 \, ^\mu A_\mu = - \beta \gamma A_0 + \gamma A_1$$
$$A_2 \rightarrow \Lambda_2 \, ^\mu A_\mu = A_2$$
$$A_2 \rightarrow \Lambda_3 \, ^\mu A_\mu = A_3$$
Decompose $\vec A = \vec A_\parallel + \vec A_\perp$ and let $\vec A_\parallel \parallel \vec \beta$, so that $$\vec A_\parallel = (A_1,0,0), \; \vec A_\perp = (0,A_2,A_2).$$ Now the Lorentz transformation is linear i.e. $$\vec A' = \vec A'_\parallel + \vec A'_\perp$$ and we can see from the components above that $\vec A_\parallel$ transforms as $$\vec A_\parallel' = \gamma(\vec A_\parallel - \vec \beta \phi)$$ and that $\vec A_\perp$ is unchanged i.e. $$\vec A'_\perp = \vec A_\perp.$$ Putting these together and rearranging, we obtain $$\vec A' =  \gamma(\vec A_\parallel - \vec \beta \phi) + \vec A_\perp = \vec A + (\gamma-1) \vec A_\parallel - \gamma \vec \beta \phi$$ it remains to show the equivalence of the middle term to that given in the solution, which is easier to do in reverse: $$\frac{\gamma^2}{1+\gamma} (\vec A \cdot \vec \beta) \vec \beta = \frac{\gamma^2 (1-\gamma)}{1-\gamma^2} A_\parallel \beta \vec \beta = \frac{1-\gamma}{1 - \beta^2 - 1} \vec A_\parallel \beta^2 = (\gamma-1) \vec A_\parallel$$ hence we can now write $$\vec A' = \vec A + \frac{\gamma^2}{1+\gamma} (\vec A \cdot \vec \beta) \vec \beta - \gamma \vec \beta \phi$$ as required.


Analogous results are obtained for the $\vec E$ and $\vec B$ fields, by considering the transformations for perpendicular components (as obtained in lectures): $$\vec E'_\perp = \gamma ( \vec E_\perp - \vec \beta \times \vec B), \; \vec B'_\perp = \gamma ( \vec B_\perp - \vec \beta \times \vec E )$$ where the extra term arises as thus: $$\vec E' = \vec E'_\perp + \vec E'_\parallel = \gamma (\vec E_\perp + \vec \beta \times \vec B) + \vec E_\parallel = \gamma (\vec E + \vec \beta \times \vec B) + (1-\gamma) \vec E_\parallel$$ and is easily cleaned up to the desired result using the procedure above. The result for $\vec B$ then follows from the duality transformation $\vec E \rightarrow \vec B, \; \vec B \rightarrow - \vec E$ with no extra calculation required.


Since $A_\mu$ is a 4-vector, then $$A_\mu A^\mu = \phi^2 - \vec A^2$$ must be a Lorentz scalar. Similarly the field strength tensors $F_{\mu \nu}$, $F^\star_{\mu\nu}$ transform according to the Lorentz transformation, and the other two invariants were shown to follow from these in lectures.

To show these results explicitly from the above transformation

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